Help with homework- 3bit, Multiplex

Discussion in 'Homework Help' started by TomS, May 29, 2011.

  1. TomS

    Thread Starter New Member

    May 29, 2011
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    Hello all,

    My name is Tom and im student of 1. year on Faculty of Electrical Engineering and Computing in Zagreb, Croatia. Since i went to Economics high school im kinda lost in my Digital electronics class, so i need some help with my homework.
    Here it goes,

    Problem:
    Using multiplexer m = 2 address inputs, realize a minimal circuit that among 3-bit complexes with 1 detects larger and equal 2, and less and equal 4.

    Thing is that i dont know how and where to start with it. Any help would be appreciated. Thanks in advance.

    p.s Sorry for english,
     
  2. Georacer

    Moderator

    Nov 25, 2009
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    Can you re-write what your circuit is supposed to do? I can't understand what you are trying to say. Write the problem description with small phrases.

    Thank you.
     
  3. TomS

    Thread Starter New Member

    May 29, 2011
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    0
    Hey, sorry about that, there are some words in croatian that is hard to translate.

    Using multiplexer m = 2 address inputs, realize a minimal circuit that among 3-bit 'combinations' with 1 detects 'combinations' which are larger and equal 2, and less and equal 4. (2<=a<=4)
     
    Last edited: May 30, 2011
  4. Georacer

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    Nov 25, 2009
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  5. TomS

    Thread Starter New Member

    May 29, 2011
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    yes, thats it. I Checked your post but i dont understand most of it. do i need to use bcd for creating mux or can i make truth table and put 1 as outputs on 2,3,4?
     
  6. Georacer

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    Nov 25, 2009
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    Last edited: May 30, 2011
  7. TomS

    Thread Starter New Member

    May 29, 2011
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    truth table:

    ABC Y
    000 0
    001 0
    010 1
    011 1
    100 1
    101 0
    110 0
    111 0

    after using k-table i got
    i0=0 -> 'a'b=0
    i1='ab'c+'abc -> 'ab='ab'c+'abc
    i2=0 -> ab=0
    i3='c -> a'b='c


    is that correct way to do it?
     
  8. Georacer

    Moderator

    Nov 25, 2009
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    Not exactly.

    The way to think about this one is to notice that the truth table has two lines for each AB combination. Four AB combinations select one out of four MUX input pins. The third column only in combination with the output Y will determine what will be connected to those input pins.

    For example:
    Look at each pair of lines in the truth table as I describe them:
    When {AB}={00}, Y=0 (in both first two lines). As a result, pin 00 (or D0 as you annotated it) will be connected to '0'.
    When {AB}={01}, Y=1. As a result, pin 01 will be connected to '1'.
    When {AB}={10}, Y=C'. Can you see that? You could make another truth table to find that but I think it is quite easy to notice. As a result, pin 10 will be connected to C'.
    Finally, pin 11 goes to 0.

    Let's recap: The pair AB will determine in which pair of lines in the truth table we are and the corresponding input pin must be connected in a signal that will produce the desired Y, in accordance to C.

    Is that clear?
     
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